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A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an α-particle to describe a circle of same radius in the same field?
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- 2 MeV
- 1 MeV
- 0.5 MeV
- 4 MeV
Correct Option: B
According to the principal of circular motion in a magnetic field
| Fc = Fm ⇒ | qVB | |
| R |
| ⇒ R = | = | = | |||
| qB | qB | qB |
| Fα = | ||
| 2qB |
| = √ | |||
| Rα | K' |
but R = Rα (given)
Thus K = K′ = 1 MeV
